So, I have a question that asks me to find the probability $P(Y<2)$ if $Y$ has a moment generating function $$M_Y(t) = (1-p+pe^t)^5$$

Is this a special distribution? Is there a trick I'm missing? Solving it algebraically/ with calculus gets really messy


Suppose $$M_X(t) = 1-p+p\exp(t)=(1-p)\exp(0t)+p\exp(1t)$$

Hence $X$ is a Bernoulli distribution with success probability $p$.

It is raised to the power of $5$, $M_Y(t)=\prod_{i=1}^5M_{X_i}(t)=M_{\sum_{i=1}^5X_i}(t)$ means $5$ independent Bernoulli trials are sum up, Hence $Y$ is a binomial distirbution with $5$ trials and success probabiity $p$.


Method 1: Genarally analysing MGF

we have $$M_Y(t)= (1-p+pe^t)^5$$ if we put 1-p=q or supposing p+q=1 we will get'


we know that binomial random variable have MGF $$M_Y(t)=(q+pe^t)^n$$ after matching the corresponding terms with our give MGF we get n=5 hence

$$Y\sim Bin (5,p)$$ so $$P(Y<2)=P(Y=0)+P(Y=1)$$ $$\Rightarrow P(Y<2)= \binom{5}{0}p^0(1-p)^5+\binom{5}{1}p^1(1-p)^4$$

$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$

Now put your value of p (i.e generally called probability of success) and get your answer

Method 2 : By generating function method

we know that

$$ M_Y(log_e(t))= G_Y(t) $$

where $M_Y(\bullet)$ denotes Moment generating function of Y and $G_Y(\bullet)$ represents generating function of Y, So we have to generally replace $t$ by $log_e(t)$ by doing that with the MGF you have given we will get


$$G_Y(t)=(1-p +pt)^5$$ $$G_Y(t)=p^5 t^5 - 5 p^5 t^4 + 10 p^5 t^3 - 10 p^5 t^2 + 5 p^5 t - p^5 + 5 p^4 t^4 - 20 p^4 t^3 + 30 p^4 t^2 - 20 p^4 t + 5 p^4 + 10 p^3 t^3 - 30 p^3 t^2 + 30 p^3 t - 10 p^3 + 10 p^2 t^2 - 20 p^2 t + 10 p^2 + 5 p t - 5 p + 1$$

as we know that $$G_Y(t)=p_0+p_1t+p_2t^2\ldots\ldots\ldots\ldots$$


$$p_0 = P(Y=0)$$

$$p_1= P(Y=1)$$




as in our Generating function we can see that $$p_0= - p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1=P(Y=0)= constant \;term \;in\;G_Y(t)$$

$$p_1= 5 p^5- 20 p^4 +30 p^3- 20 p^2+5p =P(Y=1)= coefficient\; of\; t\; in\; G_Y(t)$$

so we have to find $$P(Y<2)=P(Y=0)+P(Y=1)$$ which is $$P(Y<2)=p_0+p_1$$ $$\Rightarrow P(Y<2)=(- p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1)+(5 p^5- 20 p^4 +30 p^3- 20 p^2+5p)$$

$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$

Here method 2 is lengthy but you should keep this method in mind because in some problems it is not lengthy and it is easy , here another thing that you should notice that t have not any negative powers it means probability of negative do not exist here but in some problems where t have any negative powers there probability for negative exists


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.